## Proportion Of Ones

#### Problem

If we consider 2-digit numbers, there are exactly 18 numbers that contain the digit one:

10, 11, 12, 13, 14, 15, 16, 17, 18, 19,

21, 31, 41, 51, 61, 71, 81, 91

As there are ninety 2-digit numbers, the probability of a 2-digit number containing a one is 18/90 = 1/5.

What proportion of 3-digit numbers contain the digit one?

#### Solution

There are three mutually exclusive cases for consideration:

1**, (1')1*, and (1')(1')1.

First Digit | Second Digit | Third Digit | Combinations |

1 | 0-9 | 0-9 | 11010 = 100 |

2-9 | 1 | 0-9 | 8110 = 80 |

2-9 | 2-9 + 0 | 1 | 891 = 72 |

Total | 252 |

There are 91010 = 900 numbers that contain three digits.

Hence the proportion of 3-digit numbers containinig the digit one is 252/900 = 7/25.

Would you expect the proportion of 4-digit numbers containing the digit one to be more or less?

What about `n`-digit numbers?

Problem ID: 189 (28 Nov 2004) Difficulty: 2 Star